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A145236
a(n) is the least positive integer such that if p_n is the n-th prime then (ceiling(sqrt(a(n)*p_n)))^2 - a(n)*p_n is a perfect square.
7
2, 1, 1, 3, 5, 5, 9, 9, 13, 17, 19, 23, 25, 27, 31, 35, 41, 41, 47, 51, 51, 57, 61, 65, 73, 75, 77, 81, 83, 85, 99, 101, 107, 109, 117, 119, 125, 129, 133, 139, 145, 145, 155, 157, 161, 163, 173, 183, 187, 189, 193, 199, 201, 209, 215, 221, 225, 227, 233, 237, 239, 247
OFFSET
1,1
COMMENTS
Conjectures: 1) for n >= 2, the sequence does not decrease; 2) for n > 1, a(n) is odd; 3) a(n) can be equal to a(n+1) only for twins: p_(n+1) - p_n = 2 (although there also exist twins for which a(n) < a(n+1)).
All these conjectures are proved using the formula a(n) = p_n - 2*floor(sqrt(2p_n)) + 2, n > 1. See also A145701 and A145714. - Vladimir Shevelev, Oct 18 2008
MAPLE
A145236 := proc(n) local p, k, a ; p := ithprime(n) ; for k from 1 do ceil(sqrt(ceil(k*p))) ; a := %^2-k*p ; if issqr(a) then return k ; end if; end do: end proc:
for n from 1 do printf("%d, \n", A145236(n)) ; end do: # R. J. Mathar, Aug 02 2010
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Oct 05 2008, Oct 07 2008
EXTENSIONS
a(12)=23 (not 21). - Vladimir Shevelev, Oct 16 2008
Extended by R. J. Mathar, Aug 02 2010
STATUS
approved